# Does this infinite geometric series diverge or converge?

I am given the infinite geometric series $\sum_{n=2}^{\infty}\left(5^{n-1}\right)\left(\frac{9}{10}\right)^n$ and told that it diverges. However, my calculations indicate otherwise.

My calculations are as follows.

$\sum_{n=2}^{\infty}\left(5^{n-1}\right)\left(\frac{9}{10}\right)^n$

$= \sum_{n=2}^{\infty}\left(5^{-1}\right)\left(5^{n}\right)\left(\frac{9}{10}\right)^n$

$= \sum_{n=2}^{\infty}\left(5^{-1}\right)\left(\frac{9}{2}\right)^n$

$\dfrac{a}{1-r}$ is the formula for the sum of an infinite geometric series.

$\therefore \dfrac{a}{1-r} = \dfrac{5^{-1}}{1 - \frac{9}{2}}$

$= \dfrac{\frac{1}{5}}{\frac{-7}{2}}$

$= \dfrac{-2}{35}$

Therefore, the infinite geometric series converges to $\dfrac{-2}{35}$.

Please check my reasoning and specify where I am incorrect, why I am incorrect, and the correct reasoning to arrive at the correct answer.

Thank you.

• $\frac92>1$ so the series cannot possibly converge. The formula for the sum of a geometric series is valid when $|r|<1$, where $r$ is the ratio, which here is $\frac92>1$. – Brian M. Scott Nov 6 '16 at 23:31
• You should be suspicious when a sum of positive terms gives a negative answer. There's a condition on $r$ that must be met before you can use the formula for the sum of a geometric series. – Ethan Bolker Nov 6 '16 at 23:33
• @BrianM.Scott I see what you mean. We have $= \sum_{n=2}^{\infty}\left(5^{-1}\right)\left(\frac{9}{2}\right)^n$, where $n$ goes to infinity. Therefore, it must diverge. So under what situations is the formula $\dfrac{a}{1-r}$ invalid? – The Pointer Nov 6 '16 at 23:33
• @ThePointer: Exactly when $|r|<1$. – Brian M. Scott Nov 6 '16 at 23:34
• @BrianM.Scott I see. My thanks to both of you. Please leave your comments undeleted so that I may use them for future reference. – The Pointer Nov 6 '16 at 23:35

## 2 Answers

If we apply your reasoning, $$\sum_{n=1}^\infty 2^n=\frac{2}{1-2}=-2.$$ You should ask yourself how you get a negative result by adding all positive terms.

The reason is that the formula for the geometric series $\sum_n r^n$ applies when the series is convergent, which requires $|r|<1$.

On another note, the formula for your series (had it been convergent) would have been $$\sum_{n=2}^\infty ar^n=\frac{ar^2}{1-r}.$$

• Why $$\sum_{n=2}^\infty ar^n=\frac{ar^2}{1-r}$$? I am told that the formula for a convergent infinite geometric series is $\dfrac{a}{1-r}$. – The Pointer Nov 6 '16 at 23:37
• @ThePointer The formula you used is for a sum starting from $n=0$ but your sum starts from $n=2$. It beings 2 terms later so every term is $r^2$ times smaller. – Hugh Nov 6 '16 at 23:40
• @ThePointer: so you think that $$\sum_{n=0}^\infty ar^n=\sum_{n=1}^\infty ar^n=\sum_{n=2}^\infty ar^n.$$ – Martin Argerami Nov 6 '16 at 23:47
• @MartinArgerami You're correct. My apologies. Please leave your comment undeleted so that I may reference it in the future. – The Pointer Nov 6 '16 at 23:50

The formula you used is only suitable when $|r| < 1$. If that isn't enough, consider that a series only converges if $a_n \rightarrow 0$ as $n \rightarrow \infty$. If $|r| > 1$ then $a_n$ does not go to 0.